Question

Solve the following difference equation $\Delta\lambda^k=-k+5; \lambda^6=0$

Answer 1

Solution

Rewrite the equation in factorial powers;

∆$\lambda^k$ =-k⁽¹⁾+5

Apply the anti-difference operator ∆^-1 as;

∆^-1k⁽ⁿ⁾=$\frac{1}{(n+1)}$ kⁿ⁺¹+c

Which gives ;

$\lambda^k$ =-($\frac 12$ )k⁽²⁾+5k⁽¹⁾+c

Now we apply the given condition;$\lambda^6$=0

0=-($\frac12$)6²+5(6)+c

c=-12

So that;

$\lambda^k$=-($\frac12$)k⁽²⁾+5k⁽¹⁾-12

Convert the equation to ordinary powers of k;

We know,k⁽¹⁾=k and k⁽²⁾=-k+k²

Replace in the equation;

$\lambda^k$ =-$\frac12$(-k+k²)+5k-12

Simply;

Answer

$\lambda^k$ =-$\frac12$k²+$\frac{11}2$k-12